The number of solution of the equation
$x^{3} + x^{2} + 3 x + 2 s i n x = 0$ in $- 2 π \leq x \leq 4 π$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B 1
$2 sin x = - x (x^{2} + x + 4)$
There is no value of $x$ in between $0$ to $π$ which makes $sin x$ negative,as R.H.S is negative in $0$ to $π$
$∴$ Number of solutions of the equation $= 1$

Suggest Corrections

Similar questions

The number of solutions of the equation $x^{3}$ + $x^{2}$ + 4x + 2sinx = 0 in 0 ≤ x ≤ 2 $π$

sin 2 x + 2 sin x - cos x - 1 = 0

0 \leq x \leq 2 π

x^{3} + x^{2} + 2 x + sin x = 0

(- 2 π, 2 π)

0 \leq x \leq 2 π

2 {cos}^{2} x - 2 \sqrt{2} cos x + {tan}^{2} x - 2 tan x + 2 = 0

x \in [- 4 π, 4 π]

Join BYJU'S Learning Program